Hi All please can i have some help with this!
$\displaystyle
\int {\frac{{e^{\tan ^{ - 1} 3x} }}
{{1 - 9x^2 }}}
$
Hello, MexicanGringo!
You're answer is correct!
$\displaystyle \int \frac{e^{\arctan3x}}{1+9x^2}\,dx $
We have: .$\displaystyle \int e^{\arctan3x}\cdot\frac{dx}{1+9x^2} $
Let: $\displaystyle u \:=\:\arctan3x \quad\Rightarrow\quad du \:=\: \frac{3\,dx}{1 + 9x^2} \quad\Rightarrow\quad \frac{dx}{1+9x^2} \:=\:\frac{1}{3}\,du$
Substitute: .$\displaystyle \int e^u \cdot \frac{1}{3}\,du \;=\;\frac{1}{3}\int e^u\,du \;=\;\frac{1}{3}\,e^u + C$
Back-substitute: .$\displaystyle \frac{1}{3}\,e^{\arctan3x} + C$